Another way of looking at the solution to the problem: -

The number of ways in which m+n+p+q things can be divided into 4 groups containing m,n,p,q things severally is (m+n+p+q)!/(m!*n!*p!)

When m=n=p=q, then the formula takes the form (4m)!/(m!*m!*m!*m!)

But this formula regards as different all the possible orders in which the 4 groups can occur in any one mode of sub-division. And since there are 4! such orders corresponding to each sub-division, the number of different ways in which subdivision into 4 equal groups can be made is

(4m)!/(m!*m!*m!*m!*4!)

So, applying this to the problem we have m = 2

and (4*2)!/(2!*2!*2!*2!*4!) = 40320/384 = 105

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Mayur